Let S be a non empty set in $mathbb{R}^n$ Then, the polar cone of S denoted by $S^*$ is given by $S^*=left {p in mathbb{R}^n, p^Tx leq 0 : forall x in S right }$.

Remark

• Polar cone is always convex even if S is not convex.

• If S is empty set, $S^*=mathbb{R}^n$.

• Polarity may be seen as a generalisation of orthogonality.

Let $Csubseteq mathbb{R}^n$ then the orthogonal space of C, denoted by $C^perp =left { y in mathbb{R}^n:left langle x,y right rangle=0 forall x in C right }$.

Lemma

Let $S,S_1$ and $S_2$ be non empty sets in $mathbb{R}^n$ then the following statements are true −

• $S^*$ is a closed convex cone.

• $S subseteq S^{**}$ where $S^{**}$ is a polar cone of $S^*$.

• $S_1 subseteq S_2 Rightarrow S_{2}^{*} subseteq S_{1}^{*}$.

Proof

Step 1 − $S^*=left { p in mathbb{R}^n,p^Txleq 0 : forall 😡 in S right }$

• Let $x_1,x_2 in S^*Rightarrow x_{1}^{T}xleq 0 $ and $x_{2}^{T}x leq 0,forall x in S$

  For $lambda in left ( 0, 1 right ),left [ lambda x_1+left ( 1-lambda right )x_2 right ]^Tx=left [ left ( lambda x_1 right )^T+ left {left ( 1-lambda right )x_{2} right }^{T}right ]x, forall x in S$

  $=left [ lambda x_{1}^{T} +left ( 1-lambda right )x_{2}^{T}right ]x=lambda x_{1}^{T}x+left ( 1-lambda right )x_{2}^{T}leq 0$

  Thus $lambda x_1+left ( 1-lambda right )x_{2} in S^*$

  Therefore $S^*$ is a convex set.

• For $lambda geq 0,p^{T}x leq 0, forall 😡 in S$

  Therefore, $lambda p^T x leq 0,$

  $Rightarrow left ( lambda p right )^T x leq 0$

  $Rightarrow lambda p in S^*$

  Thus, $S^*$ is a cone.

• To show $S^*$ is closed, i.e., to show if $p_n rightarrow p$ as $n rightarrow infty$, then $p in S^*$

  $forall x in S, p_{n}^{T}x-p^T x=left ( p_n-p right )^T x$

  As $p_n rightarrow p$ as $n rightarrow infty Rightarrow left ( p_n rightarrow p right )rightarrow 0$

  Therefore $p_{n}^{T}x rightarrow p^{T}x$. But $p_{n}^{T}x leq 0, : forall x in S$

  Thus, $p^Tx leq 0, forall x in S$

  $Rightarrow p in S^*$

  Hence, $S^*$ is closed.

Step 2 − $S^{**}=left { q in mathbb{R}^n:q^T p leq 0, forall p in S^*right }$

Let $x in S$, then $ forall p in S^*, p^T x leq 0 Rightarrow x^Tp leq 0 Rightarrow x in S^{**}$

Thus, $S subseteq S^{**}$

Step 3 − $S_2^*=left { p in mathbb{R}^n:p^Txleq 0, forall x in S_2 right }$

Since $S_1 subseteq S_2 Rightarrow forall x in S_2 Rightarrow forall x in S_1$

Therefore, if $hat{p} in S_2^*, $then $hat{p}^Tx leq 0,forall x in S_2$

$Rightarrow hat{p}^Txleq 0, forall x in S_1$

$Rightarrow hat{p}^T in S_1^*$

$Rightarrow S_2^* subseteq S_1^*$

Theorem

Let C be a non empty closed convex cone, then $C=C^**$

Proof

$C=C^{**}$ by previous lemma.

To prove : $x in C^{**} subseteq C$

Let $x in C^{**}$ and let $x notin C$

Then by fundamental separation theorem, there exists a vector $p neq 0$ and a scalar $alpha$ such that $p^Ty leq alpha, forall y in C$

Therefore, $p^Tx > alpha$

But since $left ( y=0 right ) in C$ and $p^Tyleq alpha, forall y in C Rightarrow alphageq 0$ and $p^Tx>0$

If $p notin C^*$, then there exists some $bar{y} in C$ such that $p^T bar{y}>0$ and $p^Tleft ( lambda bar{y} right )$ can be made arbitrarily large by taking $lambda$ sufficiently large.

This contradicts with the fact that $p^Ty leq alpha, forall y in C$

Therefore,$p in C^*$

Since $x in C^*=left { q:q^Tpleq 0, forall p in C^* right }$

Therefore, $x^Tp leq 0 Rightarrow p^Tx leq 0$

But $p^Tx> alpha$

Thus is contardiction.

Thus, $x in C$

Hence $C=C^{**}$.